ZX Review #5-6 04 ноября 1997

Программирование     TR-DOS

forum - Reduction in the time format. On the recording sectors while formatting. Rebuilding the screen for one interrupt.

(C) M. Kuzmin, Moscow Region.


   REVIEW In an article about Roshchina
reduction in the time format. The above is there a way 
theoretically good, the practice worse: on different drives 
while formatting a single track different, and therefore bind 
to the time zero can not be done. But in general, as can be 
done during the interruption formatting, if the program

will hang in DOS'e on the block
record??

   Writing sectors simultaneously
with formatting - just a waste of time. It lost to scan sectors 
to format (To detect # F5, # F6, # F7).

Yes, of course, if these sectors do not have to track, then the 
payoff will be, but if any one sector "Unclean" (especially 
from the past on the track), it still lost the turnover drive 
on his record - loss is clearly greater.



   Ca. Ed.: Indeed, if the formatting a track by means of 
TR-DOS, you can not interrupt it. Hence, to directly control 
the VG93. An example of an algorithm format: 

 - Written in the instruction register code to VG93
   Manda formatting;
 - Grant to SH bytes corresponding

   selected format, as long as

   will be formatted in the last sector;
 - Use the force

   break;


   As can be seen, while formatting a
track does not play a role, because it is generally not used. 
But it turns out one unpleasant fact: not enough speed Z80 in 
order to give on SH bytes to format. Indeed, the disc rotates 
at a speed of 300 rpm for 1 turnover is spent 0.2 seconds. or

700,000 cycles Z80. Since the track is recorded about 7000 
bytes to write each of them must be spent not

more than 100 cycles Z80. Meanwhile, the fragment
program that writes to the disk a
bytes will be performed at least 120 cycles:
140.
; Let the registers have the following meanings:
; C - # 7F (register address data)
; HL - # 20AF (address n / n write bytes)
; IX - address the following command to which
, It hands over control after burning
; Bytes.


        LD D, B 3, 7 clocks

        PUSH IX 3, 1915

        PUSH HL; 3 11

        JP # 3D30; 3 10
# 3D30: RET; 3 10
# 20AF: LD B, March 1, 7
# 20B1: IN A, (255); 3 11

        AND # C0; 3 7

        JR Z, # 20B1; 3 7 / 12

        RET M 3, 5 / 11

        OUT (C), D 3, 1912

        DJNZ # 20B1; 3 8

        RET; 3 10
2

   And all because it registers the SH appear in the address 
space only at the moment of TR-DOS. If there was free access to 
the registers, without the involvement of TR-DOS, then 
ultra-fast format would be a reality! 

   All of the above applies to the record
sectors simultaneously with formatting.
The check for the presence of the sector of bytes # F5, # F6 # 
F7 and should be done direct formatting, so

that the extra time it takes. Yes, if these bytes will be 
found, lost another turnover disc for recording sector. But not 
so often they occur - for example, on a disk with text files of 
their may be absent.



   Corr.: Now, about rebuilding the screen for one interrupt:
as mentioned in various sources, it is impossible, but in 
specific cases, it is a very really. Take, for example, ACTION, 
LIFE SUX - demo. There's no just rebuilt the whole screen, but 
in a grid is displayed under a rather large sprite in the 
second line runs. Done this is simple: how could

note that these screens are composed of
identical segments (for example,
in LIFE SUX 2 * 2 familiarity). Program to output a single line
is this:


     LD SP, scradr +1

     LD HL, BYTE

     PUSH HL; 16 times (16 * 2 = 32)


   As can be seen, the output of one line
takes 11 * 16 +10 +10 = 191 bar, and
as one line of video controller passes for 224 cycles,
there are still plenty of time (if
take into account also the upper and BORDER
below).


   HELP ME! I have done recently
scroller (a hobby I have is:
do different VIEWER'y, text scrollers, help'y, etc.), and
faced with a problem:
stack was repositioned on the screen
(Read / write lines), but were
included IM 2, and although the program
Interrupt rearranged in the stack buffer, the screen is still 
spoiled, because Z80 when you call IM 2 still puts on a stack 
of two bytes ... In the ZX-FORMAT'e Mednonogov said that the 
combined output Music and sprites through the stack -

can anyone help
this issue - would be very grateful ...


   Ca. Ed.: If the transfer of data through the stack and 
allowed termination, you must carefully plan time distribution 
Z80. Suppose that the interval between two interruptions 70,000 
cycles procedure interrupt handling is

10,000 cycles, and the procedure for SCR, with
stack shear screen content, is 100,000 cycles. Then it is 
better to break this procedure into several parts, say,

5 for 20000 cycles each, and the program
will be as follows:


    HALT; performed procedure for handling

          ; Interrupt - 10000 cycles

    CALL SCR1; 20000

    CALL SCR2; 20000

    CALL SCR3; 20000


    HALT; 10000

    CALL SCR4; 20000

    CALL SCR5; 20000

    CALL SCR1; 20000 (next shift)

    ...